L-7 ECE-495 595

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ECE 495/595 (aka ECE-381) Introduction to Power Systems Edward D. Graham, Jr. Ph.D., P.E. Lecture 7 September 7, 2011 Office: ECE-235B [email protected] More Three Phase Power

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ECE 495/595 (aka ECE-381)Introduction to Power SystemsEdward D. Graham, Jr. Ph.D., P.E.Lecture 7September 7, 2011Office: [email protected] Three Phase Power= instantaneous powert- iRemember: Complex Conjugate (a + jb)*= (a jb)- i0NO40RootSquareMean2PF = Power Factor= Complex PowerThis equation is the basis for the Power TriangleChanging Landscape of Power Systems and Utilities DeregulationThree Phase ExampleAssume a A-connected load is supplied from a 3|13.8 kV (L-L) source with Z = 100Z20O13.8 013.8 013.8 0abbccaV kVV kVV kV= Z = Z12 = Z12 13.8 0138 20138 140 138 0abbc cakVI ampsI amps I ampsZ = = Z 100Z20 O= Z = Z10 Three Phase Example (continued)*138 20 138 0239 50 amps239 170 amps 239 0 amps3 3 13.8 0 kV 138 amps5.7 MVA5.37 1.95 MVApf cos 20 lagginga ab cab cab abI I II IS V Ij= = Z Z10 = Z = Z = Z7 = = Z Z20= Z20= += = 0.94In the News: New CWLP GeneratorCWLP = City Water, Light & Power Springfield, Ill.This is a 280 MVA Generator for CWLPs New Coal PlantDelta-Wye TransformationYphaseTo simplify analysis of balanced 3 systems:1) -connected loads can be replaced by 1Y-connected loads with Z32) -connected sources can be replaced byY-connected sources with V3 30LineZV|A==Z Delta-Wye Transformation ProofFrom the side we getHenceab ca ab caaab caaV V V VIZ Z ZV VZIA A AAA= ==Delta Wye TransformationDelta-Wye Transformation, (Continued)aFrom the side we get( ) ( )(2 )Since I 0Hence 331Therefore3ab Y a b ca Y c aab ca Y a b cb c a b cab ca Y aab caYaYYV Z I I V Z I IV V Z I I II I I I IV V Z IV VZ ZIZ ZAA= = = + + = = == ==Three Phase Transmission LineInterconnected North American Power GridThree Phase Transmission LinePer Phase AnalysisPer phase analysis allows analysis of balanced 3| systems with the same effort as for a single phase systemBalanced 3| Theorem: For a balanced 3| system with All loads and sources Y connected No mutual Inductance between phasesPer Phase Analysis (continued)Then All neutrals are at the same potential All phases are COMPLETELY decoupled All system values are the same sequence as sources. The sequence order weve been using (phase b lags phase a and phase c lags phase a) is known as positive sequence (abc); later in the course well discuss negative and zero sequence systems. Per Phase Analysis ProcedureTo do per phase analysis1. Convert all A load/sources to equivalent Ys2. Solve phase a independent of the other phases3. Total system power S = 3 Va Ia*4. If desired, phase b and c values can be determined by inspection (i.e., 120 degree phase shifts)5. If necessary, go back to original circuit to determine line-line values or internal A values. Per Phase ExampleAssume a 3|, Y-connected generator with Van = 1Z0 volts supplies a A-connected load with ZA= -jO through a transmission line with impedance of j0.1O per phase. The load is also connected to a A-connected generator with Vab= 1Z0through a second transmission line which also has an impedance of j0.1O per phase.Find1. The load voltage Vab2. The total power supplied by each generator, SY and SAPer Phase Example (continued)First convert the delta load and source to equivalent Y values and draw just the "a" phase circuitPer Phase Example (continued)' ' 'a a aTo solve the circuit, write the KCL equation at a'1(V 1 0)( 10 ) V (3 ) (V j3j j Z + + Z 30)(10 ) = 0Per Phase Example (continued)' ' 'a a a'a' 'a b' 'c abTo solve the circuit, write the KCL equation at a'1(V 1 0)( 10 ) V (3 ) (V j310(10 60 ) V (10 3 10 )3V 0.9 volts V 0.9 volts V 0.9 volts V 1.56j jj j j j Z + + Z 30)(10 ) = 0+ Z = += Z10.9 = Z130.9= Z109.1 = Z19. volts 1Per Phase Example (continued)*'*ygen*" '"S 3 3 5.1 3.5 VA0.13 5.1 4.7 VA0.1a aa a aa agen aV VV I V jjV VS V jjA| |= = = + |\ .| |= = |\ .